Convergence of Roots at Infinity

[Adgorn]

Homework Statement

Hi everyone, I'm currently making my way through Spivak's calculus and got stuck in question 41 of chapter 5. It's important to note that at this point, the book has only reached the subject of limits (haven't reached continuous functions, derivatives, integrals, series etc.).

a) For $c>1$, show that $c^{\frac 1 n}=\sqrt[n] c$ approaches 1 as $n$ becomes very large.
b) More generally, if $c>0$, then $c^{\frac 1 n}$ approaches 1 as $n$ becomes very large.

Homework Equations

Bernoulli inequality: $(1+h)^n≥1+nh$ for $h>-1$

The Attempt at a Solution

I managed to solve a): Since $c>1$, $c^{\frac 1 n}>1$. So suppose $c^{\frac 1 n}-1>\epsilon$, in other words $c>(1+\epsilon)^n≥1+n\epsilon$, this means that $n<\frac {c-1} \epsilon$. Thus, for $n≥\frac {c-1} \epsilon$, $1<c^{\frac 1 n}<\epsilon$. This solution is the same in the answer book.

I got stuck in b) (the answer book does not contain a solution for it, neither does the end of the study book). The case is trivial for $c=1$, so all that's really left to prove is the case for $0<c<1$, which I can't seem to manage. Since $c^{\frac 1 n}<1$, I need to prove $1-c^{\frac 1 n}<\epsilon$. Trying to same method as in a), if $1-c^{\frac 1 n}>\epsilon$, then $c<(1-\epsilon)^n$. This time, the Bernoulli inequality is of little use, assuming $\epsilon<1$, all it gives it $(1-\epsilon)^n≥1-n\epsilon$, and I can't really do anything with that.

Next I tried reverse engineering, if $1-c^{\frac 1 n}<\epsilon$, then $c>(1-\epsilon)^n≥1-n\epsilon$ (assuming $\epsilon<1$), and so $n>\frac {1-c} \epsilon$. But I can't prove the opposite way (which is what I need), since assuming $n>\frac {1-c} \epsilon$ only gives $c>1-n\epsilon$, and I can't really do anything with it.

So yeah, I seem to be missing something and help would be appreciated.

 

[Orodruin]

What is $(1/c)^n$ if $0<c<1$?

 

[Adgorn]

What is $(1/c)^n$ if $0<c<1$?

When $n$ grows very large it approaches infinity, but I don't seem to be able to incorporate that into the solution. The comment gave me the idea to replace the $c>1$ in the proof of a) with $\frac 1 c>1$ so that $0<c<1$, but I reached the same dead end where I can only say how small $n$ is if I assume $\epsilon<1$...

 

[pasmith]

Has Spivak proven basic results about limits of sums, products and quotients of sequences at this point? If so, you should be able to establish the following proposition:

If $a_n$ is a sequence such that $\lim_{n \to \infty} a_n = 1$ then $$\lim_{n \to \infty} \frac 1{a_n} = 1.$$

 

[Orodruin]

When $n$ grows very large it approaches infinity, but I don't seem to be able to incorporate that into the solution. The comment gave me the idea to replace the $c>1$ in the proof of a) with $\frac 1 c>1$ so that $0<c<1$, but I reached the same dead end where I can only say how small $n$ is if I assume $\epsilon<1$...

Sorry, I meant to write $(1/c)^{1/n}$. How does this number behave as $n$ becomes large and how does it relate to $c^{1/n}$?

 

[Adgorn]

Has Spivak proven basic results about limits of sums, products and quotients of sequences at this point? If so, you should be able to establish the following proposition:

If $a_n$ is a sequence such that $\lim_{n \to \infty} a_n = 1$ then $$\lim_{n \to \infty} \frac 1{a_n} = 1.$$

At this point he has not touched sequences at all, this question is more of a sneak peak than anything else.

Sorry, I meant to write $(1/c)^{1/n}$. How does this number behave as $n$ becomes large and how does it relate to $c^{1/n}$?

Got it. If $0<c<1$, then $0<c^{\frac 1 n}<1$ and $0<1<\frac 1 c$. This means that by a), if $n>\frac {\frac 1 c-1} \epsilon>0$, then $\frac 1 {c^\frac 1 n}-1<\epsilon$, meaning that $1-c^\frac 1 n<\epsilon c^\frac 1 n<\epsilon$.
Thanks for the help.